Question

# The locus of the midpoints of chords of the circle $${ x }^{ 2 }+{ y }^{ 2 }-2x-2y-2=0$$ which makes an angle $$120^{0}$$ at the center is

A
x2+y22x2y+1=0
B
x2+y2xy+1=0
C
x2+y22x2y1=0
D
None of these

Solution

## The correct option is A $${ x }^{ 2 }+{ y }^{ 2 }-2x-2y+1=0$$Given equation of circle is $${ x }^{ 2 }+{ y }^{ 2 }-2x-2y-2=0$$Let mid point of the chord $$AB$$ is $$(h,k)$$Its center is $$(1,1)$$ and radius $$=\sqrt { 1+1+2 } =2=OB$$.In $$\triangle OPB, \angle OBP ={30}^{0}$$$$\displaystyle \therefore \sin { { 30 }^{ 0 } } =\frac { OP }{ 2 } \Rightarrow OP=1$$Since $$OP=1\Rightarrow { \left( h-1 \right) }^{ 2 }+{ \left( k-1 \right) }^{ 2 }=1$$$$\Rightarrow { h }^{ 2 }+{ k }^{ 2 }-2h-2k+1=0$$ $$\therefore$$ Locus of the mid point of chord is$${ x }^{ 2 }+{ y }^{ 2 }-2x-2y+1=0$$Maths

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